Let $X$ denote the diameter of an armored electric
cable and $Y$ denote the diameter of the ceramic
mold that makes the cable. Both $X$ and $Y$ are scaled
so that they range between $0$ and $1$. Suppose that $X$
and $Y$ have the joint density

when i solved it i got the following limits of x and y, 0->1/2 and 0->(1/2-x) however according to the book the correct limits are 0->1/4 and x->1/2-x
I am just confused how to plot this function in order to find out where that 1/4 came from ?

2 Answers
2

It all hinges on drawing the right picture. Once you do that, the rest is almost automatic.

The joint density is $0$ except on or inside the triangle with vertices $(0,0)$, $(1,1)$, and $(0,1)$. We are interested in the integral of the joint density over the part of this triangle that has $x+y \gt 1/2$. So draw the line $x+y=1/2$. We will want to be "above" this line.

Note that the line $x+y=1/2$ meets the line $y=x$ at $(1/4,1/4)$. So we want to integrate the joint density over the quadrilateral-shaped region that has the following corners: $(1/4,1/4)$, $(1,1)$, $(0,1)$ and $(0,1/2)$.

The region is slightly ugly. Whether we integrate first with respect to $x$ or with respect to $y$, we will have to break up the region.

It is tempting to integrate over the part of our big triangle that has $x+y \le 1/2$, and subtract the result from $1$. This "complementary" region is the triangle with corners $(0,0)$, $(1/4,1/4)$, and $(0,1/2)$. Why $(1/4,1/4)$? Because that's where $y=x$ and $x+y=1/2$ meet.

If we integrate first with respect to $y$, there is no need to break up the integral. For then $y$ goes from $x$ to $1/2-x$. Then we integrate with respect to $x$. The rightmost point of our region is at $x=1/4$, $y=1/4$ so we will integrate from $x=0$ to $x=1/4$. that is the book's solution.

Remark: The book's solution is not optimal. Despite the need to break up the integral, I would prefer to set things up so that I integrate first with respect to $x$. Since the density function does not mention $x$ explicitly, the first integration is trivial. We can integrate over the part of the triangle that has $x+y>1/2$, or over the complementary region. Let's find the answer directly. For $y=1/4$ to $y=1/2$, we want to integrate from $x=1/2-y$ to $x=y$. From $y=1/2$ to $y=1$, we want to integrate from $x=0$ to $x=y$. No integration of $\ln$ is needed. We get
$$\int_{1/4}^{1/2} \left(2-\frac{1}{2y}\right)dy+\int_{1/2}^{1}dy,$$
which is easy to calculate.

This is mostly a rewording of things already in Andre's answer, but still it might be helpful.

When you let $y$ run from $0$ to $(1/2)-x$, you are allowing $y\le x$, but $f(x,y)=0$ there, so if you want to wind up integrating $1/y$, you want $y$ to run from $x$ to $(1/2)-x$. Also, if you want $x+y\lt1/2$, then that, together with $0\lt x\lt y$, forces $x\lt1/4$; if $x\ge1/4$, then $y\gt x\ge1/4$, and $x+y\gt1/4+1/4=1/2$. That's where the $1/4$ comes from.